What is the Proof for (a^(-1))^(-1) = a?

[STEMucator]

Homework Statement

Trying to run through every problem i can in my book in preparation for my exam. I've solved this one before, but it slipped my mind how to do it :

http://gyazo.com/9fcf9f3cef522c3d5eb1fa7d4ad04394

Homework Equations

Working in a group, so group axioms I suppose.

The Attempt at a Solution

I forgot where to start this one off, I was thinking :

e = aa^-1
a^-1e = a^-1aa^-1

That won't get me anywhere though, any pointers would be appreciated.

 

[micromass]

Do you know that every element in a group has a unique inverse?? You can use this by showing that $a$ and $(a^{-1})^{-1}$ are inverses of the same element. So they must be equal.

 

[STEMucator]

Do you know that every element in a group has a unique inverse?? You can use this by showing that $a$ and $(a^{-1})^{-1}$ are inverses of the same element. So they must be equal.

Ah, so what you're saying is if a is a group element, then it has an inverse which is also a group element denoted by a^-1.

Since a^-1 is also a group element and the inverse of a, then (a^-1)^-1 is also a group element and is the inverse of a^-1.

 

[micromass]

Ah, so what you're saying is if a is a group element, then it has an inverse which is also a group element denoted by a^-1.

Since a^-1 is also a group element and the inverse of a, then (a^-1)^-1 is also a group element and is the inverse of a^-1.

Yes. So $a^{-1}$ has two inverses. Those inverses must equal.

 

[I like Serena]

Hi Zondrina!

In a group every element has to have an inverse.
Now suppose we have b=a^-1.
Then according to the group axioms we have: ab=e
What happens if you multiply the left and right hand sides with b^-1?